Question: $f(x) = \dfrac{ \sqrt{ 6 - x } }{ \sqrt{ x - 1 } }$ What is the domain of the real-valued function $f(x)$ ?
First, we need to consider that $f(x)$ is undefined anywhere where either radical is undefined, so anywhere where either radicand (the expression under the radical symbol) is less than zero. The top radical is undefined when $6 - x < 0$ So the top radical is undefined when $x > 6$ , so we know $x \leq 6$ The bottom radical is undefined when $x - 1 < 0$ So the bottom radical is undefined when $x < 1$ , so we know $x \geq 1$ Next, we need to consider that $f(x)$ is undefined anywhere where the denominator, $\sqrt{ x - 1 }$ , is zero. So $\sqrt{ x - 1 } \neq 0$ , so $x - 1 \neq 0$ , so $x \neq 1$ So we have three restrictions: $x \leq 6$ $x \geq 1$ , and $x \neq 1$ Combining these three restrictions, we know that $1 < x \leq 6$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid 1< x \leq6\, \}$.